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JH
Numerade Educator

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Problem 72 Hard Difficulty

Evaluate the integral.

$ \displaystyle \int \frac{\ln (x + 1)}{x^2}\ dx $

Answer

$\frac{-\ln (x+1)}{x}-\ln |x|-\ln (x+1)+C$

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Video Transcript

let's start off here with the use up and take you to be X plus one. No. So do you equals the X. And we also can write you minus one equals X. So this becomes in a girl, Ellen, You over x Claire So that you might this once Where do you So that should be a two down there in the exponents because of the square up here. Now we have a new wonder girls to work with And for this new in a grove, let's go ahead and use by parts. Let's take while amusing you already let me use w equals Alan. You dd d w equals one over you deal and then Devi will be you minus one to the minus two. So use the power rule here to integrate and you get negative one over You might this one now using the formula for my parts. In this case, it's w v rights in Abril the dw So we get negative natural log of you over you minus one and then minus and a girl and that we have V and then dw solar multiplying these together we see that there's a minus sign here. This will turn this into a plus sign. And then we just have one over you from this. You up here and then you might just want from over here to you, which is coming from here. So let me just call on coordinate those this coming from there, and then you might a swan. You mind this one? Now, we could go ahead and rewrite this instagram as a over you Beale Review minus one. That's our partial freshen the composition and then we'LL have to go ahead and find a and B. So here, let's write down a V. So it turns out that Bea is one. So we have won over you, minus one and then a cz negative. Once I pull off the minus and then I integrate. So we have a natural log. Absolute value. You might this one. The minus natural are absolute value. You plus our constancy, and then go back to the definition of you the very first part of this problem. But that's how we defied you. So go ahead and replace all the use with X plus one. So here I have X plus one, minus one. So see dick here. Natural log, Absolute value X that minus l end off X plus one. You could put the absolute value here. The reason I dropped was because there are no absolute values. And for this one, this thing, this came from integration by parts. And if this war to make sense, that this term over here was also makes sense. So if we're not going to use absolute value and the first one we mind is what That you set off the second good. So it's going to raise this scratch marks over here, and then that's our final is er.